A die is thrown until the first time the total sum of the face values of the die is 700 or greater. Estimate the probability that, for this to happen,
First, we will find the expected value and variance.
\[ E[X] = \frac{\text{sum of possible outcomes}}{\text{number of outcomes}} = \frac{1+2+3+4+5+6}{6}=\frac{7}{2} \approx 3.5\\ Var[X] = E[X^2] - (E[X])^2\\ E[X^2] = \frac{1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2}{6}= \frac{91}{6}\\ Var[X] = \frac{91}{6} - (\frac{7}{2})^2 = \frac{35}{12} \approx 2.91667\\ \]
outcomes <- c(1,2,3,4,5,6)
sum_outcomes <- sum(outcomes)
count_outcomes <- length(outcomes)
EX <- sum_outcomes / count_outcomes
print(EX)
## [1] 3.5
outcomes_sq <- c(1^2,2^2,3^2,4^2,5^2,6^2)
sum_outcomes_sq <- sum(outcomes_sq)
EX_sq <- sum_outcomes_sq/count_outcomes
varx <- EX_sq - (EX)^2
print(varx)
## [1] 2.916667
The sum of each outcome is independent and identically distributed.
\[ E[S_N] = E[\Sigma_{i=1}^{N}X[i]] = \Sigma_{i=1}^N E[X[i]] = \Sigma_{i=1}^N = \frac{7}{2}N\\ Var[S_N] = Var(\Sigma_{i=1}^N X[i]) = \Sigma_{i=1}^N Var[X[i]] = \Sigma_{i=1}^N \frac{35}{12} = \frac{35}{12}N\\ \]
\[ P(S_{210} < 700)\\ E[S_{210}] = \frac{7}{2}\cdot 210 = 735\\ Var[S_{210}] = \frac{35}{12}\cdot 210 = 612.5\\ P(S_{210}<700)= P(\frac{S_{210}-E[S_{210}]}{\sqrt{Var[S_{210}]}} < \frac{700 - E[S_{210}]}{\sqrt{Var[S_{210}]}})\\ =P(Z<\frac{700-735}{\sqrt{612.5}}) \approx P(Z<-1.414) \approx 0.0787 \]
z1 <- ((700-735)/sqrt(612.5))
x1 <- pnorm(z1)
print(x1)
## [1] 0.0786496
\[ P(S_{190}\geq700)\\ E[S_{190}] = \frac{7}{2}\cdot 190 = 665\\ Var[S_{190}] = \frac{35}{12}\cdot 190 = \frac{3325}{6} \approx 554.167\\ P(S_{190}\geq700) = P(\frac{S_{190}-E[S_{190}]}{\sqrt{Var[S_{190}]}} \geq \frac{700 - E[S_{190}]}{\sqrt{Var[S_{190}]}})\\ = P(Z \geq \frac{700-665}{\sqrt{554.167}}) = P(Z \geq 1.48678) = 1-P(Z<1.48678) \approx 0.068536 \]
z2 <- ((700-665)/sqrt(554.167))
x2 <- 1-pnorm(z2)
print(x2)
## [1] 0.06853601
\[ P(S_{180} \leq 700 \leq S_{210}) = P(S_{180}<700)- P(S_{210} < 700) \\ E[S_{180}]=\frac{7}{2}\cdot 180 = 630\\ Var[S_{180}] = \frac{35}{12} \cdot 180 = 525\\ P(S_{180}<700)= P(\frac{S_{180}-E[S_{180}]}{\sqrt{Var[S_{180}]}} < \frac{700-E[S_{180}]}{\sqrt{Var[S_{180}]}})\\ = P(Z<\frac{700-630}{\sqrt{525}}) = P(Z<3.05505) \approx 0.9988749\\ P(S_{180}\leq 700 \leq S_{210}) = P(S_{180} < 700)- P(S_{210}< 700)\\ = P(S_{180}<700) - P(S_{210} < 700))\\ = 0.9988749-0.0787 \approx 0.9202 \]
z3 <- ((700-630)/sqrt(525))
x3 <- pnorm(z3)
print(x3)
## [1] 0.9988749
x3-x1
## [1] 0.9202253